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\(\Leftrightarrow2^x\left(1+2^4\right)=544\\ \Leftrightarrow2^x\cdot17=544\\ \Leftrightarrow2^x=32=2^5\Leftrightarrow x=5\)
\(2^x+2^{x+4}=544\)
\(\Rightarrow2^x+2^x\cdot2^4=544\)
\(\Rightarrow2^x\left(1+2^4\right)=544\)
\(\Rightarrow2^x\cdot17=544\)
\(\Rightarrow2^x=32\)
mà \(2^5=32\)
Suy ra: \(x=5\)
Vậy \(x=5\) là giá trị cần tìm.
2x + 2x + 4 = 544
=> 2x . 1 + 2x . 24 = 544
=> 2x . (1 + 16) = 544
=> 2x . 17 = 544
=> 2x = 32
=> 2x = 25
=> x = 5
Vậy x = 5.
~Study well~
#Shizu
\(2^x+2^{x+4}=544\)
\(2^x.1+2^x.2^4=544\Rightarrow2^x.\left(1+2^4\right)=544\)
\(2^x.17=544\Rightarrow2^x=544:17=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
Vậy \(x=5\)
2x + 2x + 4 = 544
\(\Rightarrow\) 2x + 2x . 24 = 544
\(\Rightarrow\)2x . ( 1 + 24 ) = 544
\(\Rightarrow\)2x = 544 ÷ 17
\(\Rightarrow\)2x = 32 = 25
\(\Rightarrow\)x = 5
Vậy : x = 5
2x+2x+4=544
2x+2x.24=544
2x.(1+16)=544
2x.17 =544
2x =544:17
2x =32=25
=>x=5
Vậy x=5
hok tốt
2x + 2x+4 = 544
<=> 2x + 2x.24 = 544
<=> 2x( 1 + 24 ) = 544
<=> 2x.( 1 + 16 ) = 544
<=> 2x.17 = 544
<=> 2x = 32
<=> 2x = 25
<=> x = 5
\(a,\Leftrightarrow2^x\left(1+2^4\right)=544\\ \Leftrightarrow2^x=\dfrac{544}{17}=32=2^5\\ \Leftrightarrow x=5\\ b,\Leftrightarrow\left(\dfrac{2}{5}-3x\right)^2=\dfrac{9}{25}\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}-3x=\dfrac{3}{5}\\3x-\dfrac{2}{5}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-\dfrac{1}{5}\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(2^x+2^{x+4}=544\\\Rightarrow2^x\cdot1+2^x\cdot2^4=544\\\Rightarrow2^x\cdot(1+2^4)=544\\\Rightarrow2^x\cdot(1+16)=544\\\Rightarrow2^x\cdot17=544\\\Rightarrow2^x=544:17\\\Rightarrow2^x=32\\\Rightarrow2^x=2^5\\\Rightarrow x=5\)
\(2^x+2^{x+4}=544\)
=>\(2^x+2^x\cdot2^4=544\)
=>\(2^x\left(1+2^4\right)=544\)
=>\(2^x\cdot17=544\)
=>\(2^x=\dfrac{544}{17}=32\)
=>\(2^x=2^5\)
=>x=5